Transcript Equations and Moles

Chemical Equations
p.13 Q1 Write balanced equations including state symbols for the
following:
(a)
zinc + oxygen → zinc oxide
2 Zn (s) + O2 (g) → 2 ZnO (s)
(b)
potassium + water
2 K (s)
(c)
→ potassium hydroxide + hydrogen
+ 2 H2O (l) → 2 KOH (aq)
+ H2 (g)
calcium carbonate → calcium oxide + carbon dioxide
CaCO3 (s)
→ CaO (s) + CO2 (g)
Ionic Equations
• These show what happens which ions actually react.
– spectator ions do not change and are omitted. They just
maintain electrical neutrality.
• Start with the balanced full formula equation.
• Covalent substances are written normally in the ionic
equation.
• Ionic substances which are in solution or molten have
their ions written out separately.
• Cancel out any ions appearing on both sides of the
equation unchanged
Ionic Equations
2 K (s)
+ 2 H2O (l) → 2 KOH (aq)
+ H2 (g)
2 K (s)
+ 2 H2O (l) → 2 K+ (aq) + 2 OH- (aq) + H2 (g)
Fe2(SO4)3 (aq) + 3 Zn (s) → 3 ZnSO4 (aq) + 2 Fe (s)
2 Fe3+ (aq) + 3 SO42- (aq) + 3 Zn (s) →
3 Zn2+ (aq) + 3 SO42- (aq) + 2 Fe (s)
2 Fe3+ (aq) + 3 Zn (s) → 3 Zn2+ (aq) + 2 Fe (s)
Fe2(SO4)3 (aq) + 6 KOH (aq) → 2 Fe(OH)3 (s) + 3 K2SO4 (aq)
2 Fe3+ (aq) + 3 SO42- (aq) + 6 K+ (aq) + 36 OH- (aq) →
2 Fe(OH)3 (s) + 6 K+ (aq) + 3 SO42- (aq)
Fe3+ (aq) + 3 OH- (aq) → Fe(OH)3 (s)
Ionic Equations for precipitation reactions
• Start by writing the formula of the
precipitate with its (s) state symbol on the
product (right hand) side.
• Write the 2 reactant ions needed to form
this product on the left side of the
equations with their (aq) state symbols.
p. 23 Q. 2
Al2O3 (s) + 6 HI (g) → 2 AlI3 (aq) + 3 H2O (l)
What masses of aluminium oxide and hydrogen iodide would be required
to produce 102 g aluminium iodide?
R.F.M. of AlI3 = 27 + (3 × 126.9) = 407.7
Moles of AlI3 = 102 / 407.7 = 0.250
Moles Al2O3 = 0.250 / 2 = 0.125
R.F.M. of Al2O3 = (2 × 27) + (3 × 16) =102
Mass of Al2O3 = 0.125 × 102 = 12.8 g
Moles of HI = 0.250 × 3 = 0.750
R.F.M of HI = 127.9
Mass of HI = 0.750 × 127.9 = 95.9 g
p. 23 Q.3
The substance ATP is important for transporting energy in living cells. a sample of
1.6270 g of ATP was analysed and found to contain 0.3853 g of carbon, 0.05178 g of
hydrogen, 0.2247 g nitrogen and 0.2981 g of phosphorus. The rest was oxygen. The
relative molecular mass of ATP is 507. What are its empirical and molecular formulae?
Oxygen mass= 1.6270 - 0.3853 – 0.05178 – 0.2247 – 0.2981 = 0.6671 g
C
:
H
:
N :
P
:
O
(0.3853/12):(0.05178/1):(0.2247/14):(0.2981/31):(0.6671/16)
=
= 0.03211 : 0.05178 : 0.01605 : 0.009616 : 0.04169
0.009616 0.009616 0.009616 0.009616
0.009616
= 3.339 :
5.385
: 1.669 :
1.000 :
4.335
=10
16
: 5
3
13
:
:
:
empirical formula = C10H16N5P3O13 empirical formula mass = 507
molecular formula mass = 1 × empirical formula mass so molecular formula =
C10H16N5P3O13
p.27 Q. 1
A stock solution contains 10 mol dm-3 of hydrochloric acid. What
volume of this solution should be dissolved in water to make 6 dm3 of
acid of concentration 0.1 mol dm-3?
Moles of HCl = 6 dm3 × 0.1 mol dm-3 = 0.6
Volume = 0.6 mole / 10 mol dm-3 = 0.06 dm3 = 60 cm3.
p.27 Q. 2
When solutions of potassium chloride and silver nitrate are mixed together, a
precipitate of silver chloride forms.
(a) Write a balanced molecular equation for this reaction.
KCl (aq) + AgNO3 (aq) → AgCl (s) + KCl (aq)
(b) In an experiment, 24.0 cm3 of 0.05 M silver nitrate solution exactly reacts
with 15.0 cm3 of potassium chloride solution. Calculate the molar
concentration of the potassium chloride.
Moles of AgNO3 = 24.0/1000 dm3 × 0.05 mol dm-3 = 0.0012
1:1 AgNO3:KCl so 0.0012 mol KCl
Concentration = 0.0012 mol / (15/1000) dm3 = 0.0800 mol dm-3
(c) How much KCl would be needed to produce 100 cm3 of this concentration?
Moles = 100/1000 dm3 × 0.0800 mol dm-3 = 0.008
Mass = 0.008 mol × 74.5 g mol-1 = 0.596 g
p.31 Question 1
A chemist mixed 12.0 g of phosphorus with 35.5 g of chlorine gas to
synthesise phophorus (III) chloride. The yield was 42.4 g of PCl3. the
equation for the reaction is:
2 P (s) + 3 Cl2 (g) → 2 PCl3 (l)
Moles of P = 12 / 31 = 0.3871;
Moles of Cl2 = 35.5 / 71 =0.5
Moles of PCl3 = 42.4 / 137.5 = 0.3084.
This amount of PCl3 required 3/2 × 0.3084 moles Cl2 which is 0.463
moles, and 0.3084 moles P.
Therefore P was used in excess and Cl2 was limiting.
Percentage yield in terms of Cl2 = (0.463 / 0.5)*100 = 92.6%
p.31 Q2
2-bromo-2-methylpropane, C4H9Br can be used to produce
methylpropene, C4H8. 2-bromo-2-methyl propane reacts with sodium
ethoxide, NaOC2H5 to produce ethanol C2H5OH and sodium bromide,
along with methylpropene. Work out the atom economy for the reaction.
Ratio
C4H9Br : C2H5OH : C4H8 = 1 : 1 : 1
R.F.M.
137
46
56
Atom economy = (56/[137 + 46]) × 100
= 30.6%
Volume of
1.0M KI
(aq) (cm3)
Moles of Volume of 1.0M
Moles of
KI
Pb(NO3)2 (aq) (cm3) Pb(NO3)2
Precipitate
depth (mm)
5
0.005
0.5
0.0005
5
5
0.005
1
0.0010
6
5
0.005
1.5
0.0015
8
5
0.005
2
0.0020
10
5
0.005
2.5
0.0025
12
5
0.005
3
0.0030
12
5
0.005
3.5
0.0035
12
Reaction between 0.005 mol KI (aq) with Pb(NO3)2
(aq)
14
Depth of precipitate (mm)
p.29
Q1
12
10
8
6
4
2
0
0
0.001
0.002
0.003
0.004
Mol Pb(NO3)2 (aq)
0.025 mol Pb(NO3)2 to 0.50 mol KI.
1:2 ratio
Pb(NO3)2 (aq) + 2 KI (aq) → PbI2 (aq) + 2 KNO3 (aq)
Isotopes
• An isotope is a sample of an element in
which the atoms have the same number of
neutrons as well as the same number of
neutrons.
• Isotopes are samples of the same element
in which the atoms have different numbers
of neutrons whilst having the same
number of protons.
Relative atomic mass
• R.A.M. = (weighted average mass of an atom) /
(1/12 mass of an atom of 12C)
• Relative isotopic mass, relative formula mass
and relative molecular mass are defined on the
same scale.
– Relative isotopic mass is not necessarily an integer
and identical to mass number
• R.A.M., R.F.M. and R.M.M. have no units
– ratios of masses
• Molar mass has the same numerical value as
R.F.M. but has units g mole-1 (‘the mass of 1
mole’)
Mass spectrometer
gaseous sample
+
X (g) + e-  X+ (g) + 2 enegative
plates
electron
gun (heated
filament)
variable
electromagnet
computer
+
amplifier
detector
vacuum
Mass spectrometer
• Gaseous sample needed (vaporisation by heating or
laser dye absorption)
• Vacuum so ions move freely without colliding
• Electron gun bombards sample molecule with high
energy electrons causing ionisation, e.g.
CH4 (g) + e- → CH4+ (g) + 2 e• Negative plates accelerate sample ions and select for
uniform velocity
• Variable electromagnet deflects the ions
• Detector, connected to computer, generates a graph of
ion abundance against mass / charge (a function of
electromagnetic field strength)
• Molecular ions can fragment by random breaking of
any covalent bond within the molecule. Every species
detected in the mass spectrum is a positive ion.
Calculating relative atomic mass
Bromine (50.5% 79Br, 49.5% 81Br)
RAM = 80.0
Silver (51.3% 107Ag, 48.7% 109Ag
RAM = {(51.3 × 107) + (48.7 × 109)} / 100 = 108.0
Chromium ( 4.3% 50Cr, 83.8% 52Cr, 9.6% 53Cr, 2.3% 54Cr)
RAM = {(4.3 × 50) + (83.8 × 52) + (9.6 × 53) + (2.3 × 54)} / 100
= 52.1
Tips: SUS, check answer is plausible.
p. 55 Q2
Isotope
78Kr
Relative isotopic mass
77.92
Percentage abundance %
0.12
80Kr
79.92
81.91
82.91
2.0
12
12
83.91
85.91
57
17
82Kr
83Kr
84Kr
86Kr
Notice that relative isotopic mass is not identical to mass number
(due to nuclear binding energy).
R.A.M. = {(0.12 × 77.92) + (2.0 × 79.92) + (12 × 81.91) + (12 ×
82.91) + (57 × 83.91) + (17 × 85.91)} / 100
= 83.90
(SUS)
Some other uses of mass spectrometry
• Determining 12C / 13C in plant derived material
– hence photosynthetic pathway and type of plant
• Dating rocks
• 14C dating (age of materials of biological origin)
• Isotopic composition of hormones in athletes (is the
molecule made ‘recently’ from material of
photosynthetic origin and is therefore ‘natural’ or was it
made from petroleum derived material by an organic
chemist?
• Analysis of atmospheres on other planets by
spacecraft.